多个分隔符用于相同的文件输入R.

Question

我已经找到了答案,但只发现了C或C#的内容.我意识到R的大部分是用C语言编写的,但我对它的了解并不存在.我也是R的新手.我正在使用当前的Rstudio.

我认为这与我想要的类似. 使用R中的多个分隔线有效地读取数据

我有一个csv文件,但是一个变量是一个字符串,其值由_和分隔- .我想知道是否有一个包或额外的代码在读取时执行以下操作.命令.

"1","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_ANDROID","2013-08-31 13:39:55.0","2013-10-16 13:58:00.0",0,218,4,93,1377907200000
"2","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_ANDROID","2013-08-31 13:39:55.0","2013-10-16 13:58:00.0",0,390,5,157,1377993600000
"3","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_ANDROID","2013-08-31 13:39:55.0","2013-10-16 13:58:00.0",0,376,5,193,1.37808e+12
"4","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_ANDROID","2013-08-31 13:39:55.0","2013-10-16 13:58:00.0",1,35,1,15,1377907200000
"5","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_ANDROID","2013-08-31 13:39:55.0","2013-10-16 13:58:00.0",12,11258,117,2843,1377993600000
"6","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_ANDROID","2013-08-31 13:39:55.0","2013-10-16 13:58:00.0",5,4659,56,1826,1.37808e+12
"7","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_ANDROID","2013-08-31 13:39:55.0","2013-10-16 13:58:00.0",7,7296,136,2684,1377907200000
"8","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_IOS_IPAD","2013-08-31 13:18:21.0","2013-10-16 13:58:00.0",0,4533,35,1632,1377907200000
"9","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_IOS_IPAD","2013-08-31 13:18:21.0","2013-10-16 13:58:00.0",0,421,6,161,1377993600000
"10","Client1","Name2","*Name3_Name1_KB_MobApp_M-13-44_AU_PI Likes by KB_IOS_IPAD","2013-08-31 13:18:21.0","2013-10-16 13:58:00.0",0,57,2,23,1.37808e+12

示例行:

Name    Name1   *XYZ_Name3_KB_MobApp_M-18-25_AU_PI ANDROID  2013-09-32 14:39:55.0   2013-10-16 13:58:00.0   0   218 4   93  1377907200000

因此,阅读时很容易

results <- read.delim("~/results", header=F)

但是我仍然有字符串*XYZ_Name3_KB_MobApp_M-18-25_AU_PI

期望的输出(单独_和通过-):

Name    Name1   *XYZ   Name3  KB   MobApp   M 18 25  AU  PI ANDROID 2013-09-32 14:39:55.0   2013-10-16 13:58:00.0   0   218 4   93  1377907200000

但没有拆分时间字符串.

----感谢@Henrik和@AnandaMahto提供的代码和包.----

library(splitstackshape)

# split concatenated column by `_`
df4 <- concat.split(data = df3, split.col = "V3", sep = "_", drop = TRUE)

# split the remaining concatenated part by `-`
df5 <- concat.split(data = df4, split.col = "V3_5", sep = "-", drop = TRUE)

Answer 1

我发现包中的函数在splitstackshape这种情况下很方便。

library(splitstackshape)

# split concatenated column by `_`
results2 <- concat.split(data = results, split.col = "V3", sep = "_", drop = TRUE)

# split the remaining concatenated part by `-`
results3 <- concat.split(data = results2, split.col = "V3_5", sep = "-", drop = TRUE)
results3

Answer 2

尝试这个：

# dummy data
df <- read.table(text="
Name    Name1   *XYZ_Name3_KB_MobApp_M-18-25_AU_PI ANDROID  2013-09-32 14:39:55.0   2013-10-16 13:58:00.0   0   218 4   93  1377907200000
Name    Name2   *CCC_Name3_KB_MobApp_M-18-25_AU_PI ANDROID  2013-09-32 14:39:55.0   2013-10-16 13:58:00.0   0   218 4   93  1377907200000
", as.is = TRUE)

# replace "_" to "-"
df_V3 <- gsub(pattern="_", replacement="-", df$V3, fixed = TRUE)

# strsplit, make dataframe
df_V3 <- do.call(rbind.data.frame, strsplit(df_V3, split = "-"))

# output, merge columns
output <- cbind(df[, c(1:2)],
                df_V3,
                df[, c(4:ncol(df))])

基于下面的评论，这里是另一个相关选项，但它使用read.table而不是strsplit.

splitCol <- "V3"
temp <- read.table(text = gsub("-", "_", df[, splitCol]), sep = "_")
names(temp) <- paste(splitCol, seq_along(temp), sep = "_")
cbind(df[setdiff(names(df), splitCol)], temp)